I noticed my approach differed from the expected way in problem div2 C.

I solved it greedily: if a room that was explored at time ti exists, then do not add to the answer, but update that room so that it is explored at time i. (since you revisited) Otherwise, increment the answer by one and then create a new room visited at time i.

Your resultant string can not contain similar letters. For example, if you have "aba" where 'a' appears two times, it is already impossible to construct a string where "aba" will be the most frequent string. Now, using this fact, let's build a graph on letters of our strings. If we have a string "curd", that means we need to create 3 edges {'c' -> 'u', 'u' -> 'r', 'r' -> 'd'}. If such directed graph does not contain loops, just restore strings and output them in lexicographical order

After reading this three times, it still doesn't make any sense "Let us note that projection of set of points to line move center of mass of initial set to center of mass of initial set to center of mass of projections multiset" I think it should be that "Let us not that the projection of the set of points to the line moves the mass center of the initial set to the mass center of the projected points". This is in the solution of div1D

When I got my code on Problem D accepted after the contest, it was more than 100 lines. I was thinking that I finally got accepted and was satisfied. But then when I saw another one's code: http://codeforces.com/contest/890/submission/32272491 It has just 40 lines!!! I was shocked. However, reading that piece of code made me think about the logic of my code and found that there was a lot of redundancy. Then I improved my code and it became less than 60 lines. I feel really lucky about the fact that we can read others' code on Codeforces.

A small addition to the 886C editorial, you also need to add 1 to the summation because the first room is NOT incorporated in the summation. This is because every term of the summation is a result of adding an "extra" room when a particular 'time' reappears in the array, ans since the first room is not added as "extra" room, it is therefore not a term in the summation. So the answer should be: ans = ans + 1

First addendum is the number of permutations where number n — 1 is among the first n — k — 1 numbers, all of this permutations are good. There are (n — k — 1) ways to place n — 1 and (n — 2)! ways to place every other number from 1 to n — 2.

The second addendum is the number of permutations, where n — 1 comes at the h-th place, h >= n — k. Such permutation is good if its prefix of size h is good. So when we decide which numbers appear among the first h — 1, the number of such permutations is D(h) * (n — h — 1)!, because the n — h — 1 numbers between n — 1 and n can come in any order. There are C_n-2_h-1 ways to pick h — 1 numbers from the n — 2. So the result of multiplying is D(h) * (n — h — 1)! * (n — 2)! / (h — 1)! / (n — h — 1)! = D(h) * (n — 2)! / (h — 1)!

To find the final answer let's fix where n is placed in the permutation. Let it be the i-th position in 1-indexation. Then the number of such permutation is the number of ways to pick up and place the numbers that come after the i-th position multiplied by the nomber of good permutations of size i, which is D(i). The first multiplier is C_n-1_i-1 * (n — i)! = (n — 1)! / (i — 1)! So, the answer is sum_1_n[D(i) * (n — 1)! / (i — 1)!].

Thanks, huesoss. I just want to put your explanation in easier to read format.

First addendum is the number of permutations where number (n - 1) is among the first (n - k - 1) numbers, all of this permutations are good. There are (n - k - 1) ways to place (n - 1) and (n - 2)! ways to place every other number from 1 to (n - 2).

The second addendum is the number of permutations, where (n - 1) comes at the hth place, (h ≥ n - k). Such permutation is good if its prefix of size h is good. So when we decide which numbers appear among the first (h - 1), the number of such permutations is D(h) * (n - h - 1)!, because the (n - h - 1) numbers between (n - 1) and n can come in any order. There are ways to pick (h - 1) numbers from the (n - 2).

So the result of multiplying is

To find the final answer let's fix where n is placed in the permutation. Let it be the ith position in 1-indexation. Then the number of such permutation is the number of ways to pick up and place the numbers that come after the ith position multiplied by the number of good permutations of size i, which is D(i). The first multiplier is . So, the answer is

In div 2 D, can anyone explain why we are checking that all the characters of initial strings should be presented in our answer String. I mean that if we are doing dfs for all characters, then why we need to check it separately? If we do not check it, it is giving WA on test 24.

"A string (not necessarily from this set) is called good if ALL ELEMENTS of the set are the most frequent substrings of this string." And it looks like the path found by your dfs doesn't include ALL characters (and sequence of characters) from the set. E.g. 't' whereas the string you found is "crehpuqmbao"

If the test case is just string "aba", then you have a cycle with no entry point, so you won't start dfs and figure out it was a cycle. You check that the answer is not empty, but if the input is "aba, xyz" then you will get "xyz".

It's inferred, not stated in the problem. Try the opposite statement: 1. Assume a string "aba" in the set is the most frequent. Inside it there is a substring that appears more than once: 2 "a" => then "a" appears more times than "aba" in the result string. => "aba" is no longer the most frequent string. Hence, the most frequent string (or all strings in the set) must have no duplicated substring in itself, i.e. every substring in it appears at most once. In short, any substring in the most frequent string(s) is only allowed to be inside any one string at most once.

It is high time Codeforces started caring about the variation of English used in editorials. I usually don't mind small mistakes, as long as the meaning is not completely lost somewhere inside this modified grammar/vocabulary.

Can anybody help me in understanding what does exactly Div2 C wants with all possible combination of sample's explanation? I read several times but failed to understand the description as well as samples. ~ TIA

Just do the O(n^2) dp with map. It is O(n log n log C)! Since we all know that one number will be zero after log(C) times mod (one mod will make it below that half) and we also notice that if the num was smaller that the mod we now is calculating , we must keep it unmoved. (We can use a lower_bound to keep the complexity right). So what's the time to get all these numbers to 1 using mods? certainly(n log C) and each time we use log n time to discover it. We can optimize the algorithm in this way.